0=4r^2+24r-28

Solution for 0=4r^2+24r-28 equation:

0=4r^2+24r-28

We move all terms to the left:

0-(4r^2+24r-28)=0

We add all the numbers together, and all the variables

-(4r^2+24r-28)=0

We get rid of parentheses

-4r^2-24r+28=0

a = -4; b = -24; c = +28;
Δ = b2-4ac
Δ = -242-4·(-4)·28
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-32}{2*-4}=\frac{-8}{-8}
=1
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+32}{2*-4}=\frac{56}{-8}
=-7
$